Specific heat capacity — calorimetry: Exercises — Hinglish deep dive

1.7.4 · D4 · Physics › Thermodynamics › Specific heat capacity — calorimetry

Poore note mein hum woh do tools use karte hain jo parent note mein banaye gaye hain:

Level 1 — Recognition

Recall Solution L1.1

WHAT we want: $Q$ , given $m$ , $c$ , $Δ T$ . Seedha $Q = m c Δ T$ ka use. WHY: substance sirf warm ho raha hai (koi melting/boiling nahi), isliye yeh akela formula kaafi hai. $Δ T = 45 - 15 = 30 K$ (°C mein gap waisi hi hoti hai jaise K mein.) $Q = 2.0 \times 4186 \times 30 = 251160 J \approx 251 kJ$

Recall Solution L1.2

WHAT: $C$ = poore object ko 1 K warm karne ke liye heat, toh mass andar hi hai. WHY: $C = m c$ — kilograms se multiply karke "per kg" hata diya jaata hai. $C = m c = 3.0 \times 450 = 1350 J K^{- 1}$

Recall Solution L1.3

WHAT: $w$ aisa ki $w c_{water} = m c_{iron}$ (same heat capacity). WHY: water equivalent object ko paani ke ek "utne hi aalsi" gole se replace karta hai. $w = \frac{m c}{c _{water}} = \frac{1350}{4186} = 0.3225 kg$

Level 2 — Application

Recall Solution L2.1

WHAT: paani se nikalne wali heat, toh expect karo $Q < 0$ (heat bahar). $Δ T = 30 - 90 = - 60 K$ $Q = 0.40 \times 4186 \times (- 60) = - 100464 J$ WHAT IT MEANS: minus sign kehta hai $\approx 100 kJ$ bahar gayi. Release hui heat ki magnitude hai $100464 J \approx 100.5 kJ$ .

Recall Solution L2.2

WHAT: $Q = m c Δ T$ ko $Δ T$ ke liye solve karo. WHY $m c$ se divide karo: $m c$ object ki heat capacity $C$ hai; "energy per degree" se energy divide karne par degrees milte hain. $Δ T = \frac{Q}{m c} = \frac{5000}{0.25 \times 900} = \frac{5000}{225} = 22.22 K$

Recall Solution L2.3

PREDICT: equal masses, equal $c$ ⇒ seedha average ⇒ $40$ °C. VERIFY weighted-average formula se: $T_{f} = \frac{m _{1} c _{1} T _{1} + m _{2} c _{2} T _{2}}{m _{1} c _{1} + m _{2} c _{2}} = \frac{1 \cdot 4186 \cdot 70 + 1 \cdot 4186 \cdot 10}{4186 + 4186}$ $4186$ upar aur neeche cancel ho jaata hai: $T_{f} = \frac{70 + 10}{2} = 40 °C ✓$

Level 3 — Analysis

Recall Solution L3.1

Figure — the tug-of-war:

Figure — Specific heat capacity — calorimetry

WHAT: hot metal

95 \to 22

thanda hota hai, cold water

18 \to 22

garm hota hai. WHY set them equal: insulated cup ⇒ metal ka heat loss = water ka heat gain. Paani ne gain ki heat:

Q_{w} = 0.20 \times 4186 \times (22 - 18) = 0.20 \times 4186 \times 4 = 3348.8 J

Metal ne khoi heat:

Q_{m} = 0.15 \times c_{m} \times (95 - 22) = 0.15 \times c_{m} \times 73

Q_{m} = Q_{w}

set karo:

0.15 \times 73 \times c_{m} = 3348.8 \Rightarrow c_{m} = \frac{3348.8}{10.95} = 305.8 J kg^{- 1} K^{- 1}

Recall Solution L3.2

WHY closer to 80: 0.30 kg body ka heat capacity $m c$ zyada bada hai, isliye woh "tug-of-war jeetta hai" aur $T_{f}$ ko apni temperature ki taraf kheenchta hai. $T_{f} = \frac{0.30 \cdot 4186 \cdot 80 + 0.10 \cdot 4186 \cdot 20}{0.30 \cdot 4186 + 0.10 \cdot 4186}$ $4186$ factor out karo: $T_{f} = \frac{0.30 \cdot 80 + 0.10 \cdot 20}{0.30 + 0.10} = \frac{24 + 2}{0.40} = \frac{26}{0.40} = 65 °C$ Sach mein $65$ °C, 80 ke zyada paas hai 20 ke comparison mein. ✓

Recall Solution L3.3

WHY not the plain average: equal masses hain lekin unequal $c$ — paani ka bada $c$ use zyada "pull" deta hai, isliye $T_{f}$ paani ke 25 °C ki taraf jhukta hai. Heat capacities: water $m c = 0.50 \times 4186 = 2093$ , oil $m c = 0.50 \times 2000 = 1000$ . $T_{f} = \frac{2093 \cdot 25 + 1000 \cdot 75}{2093 + 1000} = \frac{52325 + 75000}{3093} = \frac{127325}{3093} = 41.17 °C$ Note: plain average $50$ °C se neeche hai, bilkul jaisa predict kiya tha (paani neeche kheench raha hai).

Level 4 — Synthesis

Recall Solution L4.1

WHAT changed: ab water AUR cup dono 18 se 22 tak warm hote hain, isliye dono heat-gained side par aate hain. WHY: cup thermal contact mein hai — ise ignore karna absorbed heat ko undercount karta hai aur $c_{m}$ ko low bias karta hai. $Q_{gained} = (m_{w} c_{w} + m_{cu} c_{cu}) Δ T = (0.20 \cdot 4186 + 0.080 \cdot 386) (22 - 18)$ $= (837.2 + 30.88) \times 4 = 868.08 \times 4 = 3472.32 J$ Metal ne khoi heat ke barabar set karo: $0.15 \times c_{m} \times 73 = 3472.32 \Rightarrow c_{m} = \frac{3472.32}{10.95} = 317.1 J kg^{- 1} K^{- 1}$ L3.1 se compare karo ( $305.8$ ): cup include karne se answer badhta hai, kyunki paani akele se zyada heat truly absorb hui thi.

Recall Solution L4.2

WHY a general weighted average: teen bodies ke saath aur koi heat bahar nahi jaati, toh final temperature teeno starting temperatures ka $m c$ -weighted average hota hai. $T_{f} = \frac{\sum m _{i} c _{i} T _{i}}{\sum m _{i} c _{i}}$ Heat capacities:

water A: $0.20 \times 4186 = 837.2$ , at 15
water B: $0.10 \times 4186 = 418.6$ , at 60
Al: $0.050 \times 900 = 45$ , at 90 $Numerator = 837.2 \cdot 15 + 418.6 \cdot 60 + 45 \cdot 90 = 12558 + 25116 + 4050 = 41724$ $Denominator = 837.2 + 418.6 + 45 = 1300.8$ $T_{f} = \frac{41724}{1300.8} = 32.07 °C$ Check: 15 aur 90 ke beech hai ✓, aur bada thanda paani (837.2) dominate karta hai, $T_{f}$ ko 15–32 range ki taraf neeche kheenchta hai ✓.

Recall Solution L4.3

WHAT: water equivalent cup ke $C$ ko paani ki barabar mass mein badal deta hai. $w = \frac{C _{cup}}{c _{water}} = \frac{50}{4186} = 0.01194 kg$ WHY useful: ab sab kuch "water" hai, isliye heat capacities sirf mass ke saath scale hoti hain. Cup ko $0.01194$ kg paani maano jo 20 °C par hai, $0.10$ kg paani ke saath 90 °C par mix ho raha hai: $T_{f} = \frac{0.10 \cdot 90 + 0.01194 \cdot 20}{0.10 + 0.01194} = \frac{9 + 0.2388}{0.11194} = \frac{9.2388}{0.11194} = 82.53 °C$

Level 5 — Mastery

Recall Solution L5.1

WHAT: dono same shuru karte hain, $T_{1} = T_{2} = 40$ . $T_{f} = \frac{m _{1} c _{1} \cdot 40 + m _{2} c _{2} \cdot 40}{m _{1} c _{1} + m _{2} c _{2}} = 40 \cdot \frac{m _{1} c _{1} + m _{2} c _{2}}{m _{1} c _{1} + m _{2} c _{2}} = 40 °C$ Heat flow: har $Δ T = 0$ , toh $Q = m c \cdot 0 = 0$ dono ke liye — koi heat nahi chalti. Equal temperatures = pehle se thermal equilibrium; heat sirf temperature ke difference par flow karti hai.

Recall Solution L5.2

WHY intuition first: lake ki heat capacity ( $1000 \times 4186$ ) pebble ki ( $0.001 \times 800 = 0.8$ ) se kaafi zyada hai, isliye $T_{f}$ 10 °C se mushkil se hilega. $T_{f} = \frac{1000 \cdot 4186 \cdot 10 + 0.001 \cdot 800 \cdot 100}{1000 \cdot 4186 + 0.001 \cdot 800} = \frac{41860000 + 80}{4186000 + 0.8}$ $T_{f} = \frac{41860080}{4186000.8} = 10.0000... \approx 10.00 °C$ LIMIT: jab ek body ka $m c \to \infty$ , woh thermal reservoir ki tarah act karta hai — uska temperature $T_{f}$ set karta hai aur essentially fixed rehta hai. Yeh "heat bath" ki idealisation hai.

Recall Solution L5.3

WHY it's impossible: $T_{f}$ 30 aur 50 ka weighted average hai, isliye yeh unke beech hona chahiye. $55$ bahar hai $[30, 50]$ se — yeh pakka sign hai ki student ne "lost" aur "gained" swap kar diya (sign error). Sahi value (equal $m$ , equal $c$ ⇒ plain average): $T_{f} = \frac{30 + 50}{2} = 40 °C$

Recall Solution L5.4

WHAT: unknown cold mass $m$ hai. Same $c$ cancel ho jaata hai. Hot water ne khoi heat = cold water ne gain ki heat: $0.10 \times c_{w} \times (80 - 35) = m \times c_{w} \times (35 - 20)$ $c_{w}$ cancel karo: $0.10 \times 45 = m \times 15 \Rightarrow 4.5 = 15 m \Rightarrow m = 0.30 kg$ Check: cold body 3× bhaari hai, isliye $T_{f} = 35$ cold 20 ke zyada paas hai hot 80 ke comparison mein ✓.

Recall Self-test summary — sirf sab try karne ke baad reveal karo

L1: $Q = 251$ kJ; $C = 1350$ J/K; $w = 0.323$ kg. L2: released $100.5$ kJ; $Δ T = 22.2$ K; $T_{f} = 40$ °C. L3: $c_{m} = 305.8$ ; $T_{f} = 65$ °C; $T_{f} = 41.17$ °C. L4: $c_{m} = 317.1$ ; $T_{f} = 32.07$ °C; $w = 0.0119$ kg, $T_{f} = 82.53$ °C. L5: $T_{f} = 40$ , $Q = 0$ ; $T_{f} \approx 10.00$ °C; $T_{f} = 40$ °C; $m = 0.30$ kg.

Connections

Conservation of Energy — woh law jis par calorimetry tiki hai.
Heat and Internal Energy — "heat" $Q$ physically kya hoti hai.
First Law of Thermodynamics — energy bookkeeping generalized.
Latent Heat and Phase Change — jab temperature ruk jaata hai badhna.
Conduction Convection Radiation — heat actually bodies ke beech kaise travel karti hai.
Molar Heat Capacity of Gases — Cp and Cv — $c$ ka per-mole cousin.